[tex]x^2 -4x +1 = 0[/tex]
According to the Zero-Product Property, if you have the Quadratic Equation, [tex](x +a)(x +b) =0[/tex], it's roots will be [tex]x = -h[/tex] and [tex]x = -k[/tex].
If we want our roots to be [tex]2 -\sqrt 3[/tex] and [tex]\frac{1}{2 -\sqrt 3}\\[/tex], then [tex]-h = 2 -\sqrt 3[/tex] and [tex]-k = \frac{1}{2 -\sqrt 3}\\[/tex].
Solving for [tex]h[/tex]:
[tex]-h = 2 -\sqrt 3 \\ -h \cdot -1 = (2 -\sqrt 3) \cdot -1 \\ h = \sqrt 3 -2[/tex]
Solving for [tex]k[/tex]:
[tex]-k = \frac{1}{2 -\sqrt 3} \\ -k \cdot -1 = \frac{1}{2 -\sqrt 3} \cdot -1 \\ k = \frac{1}{\sqrt 3 -2}[/tex]
Writing the equation:
[tex](x +h)(x +k) = 0 \\ (x +(\sqrt 3 -2))(x +\frac{1}{\sqrt 3 -2}) = 0[/tex]
Unsurprisingly, the answer shouldn't be that because it shouldn't be that easy. By that, we have to apply the distributive property.
[tex](x +(\sqrt 3 -2))(x +\frac{1}{\sqrt 3 -2}) = 0 \\ x(x +(\sqrt 3 -2)) +\frac{1}{\sqrt 3 -2}(x +(\sqrt 3 -2)) = 0 \\ x^2 +(\sqrt 3 -2)x +\frac{1}{\sqrt 3 -2}x +\frac{1}{\sqrt 3 -2}(\sqrt 3 -2) = 0 \\ x^2 +((\sqrt 3 -2) +\frac{1}{\sqrt 3 -2})x +1 = 0 \\ x^2 + (\frac{(\sqrt 3 -2)^2}{\sqrt 3 -2} +\frac{1}{\sqrt 3 -2})x +1 = 0 \\ x^2 +\frac{3 -4\sqrt 3 +4 +1}{\sqrt 3 -2}x +1 = 0 \\ x^2 +\frac{8 -4\sqrt 3}{\sqrt 3 -2}x +1 = 0 \\ x^2 +\frac{-4(\sqrt 3 -2)}{\sqrt 3 -2}x +1 = 0 \\ x^2 -4x +1 = 0[/tex]
